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22 Apr 2010, 09:10
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Hi all!
What is the remainder when 2^91 is divided by 7?
Correct Answer: 2
Can anyone explain how to approach such questions?
Thx in advance!
What is the remainder when 2^91 is divided by 7?
Correct Answer: 2
Can anyone explain how to approach such questions?
Thx in advance!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
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22 Apr 2010, 09:46
Kudos
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there is a pattern to remainder here:
2^1 = 2/7 R=2
2^2 = 4/7 R=4
2^3 = 8/7 R=1
2^4 = 16/7 R = 2
2^5 = 32/7 R =4
etc....
So you see the pattern of remainder is 2,4,1 and it repeats every 3 powers so 3 goes into 91 30 times with 1 left over - that being the first of the pattern so R=2
2^1 = 2/7 R=2
2^2 = 4/7 R=4
2^3 = 8/7 R=1
2^4 = 16/7 R = 2
2^5 = 32/7 R =4
etc....
So you see the pattern of remainder is 2,4,1 and it repeats every 3 powers so 3 goes into 91 30 times with 1 left over - that being the first of the pattern so R=2
22 Apr 2010, 09:52
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gmatt14
the expression \(2^{91}\) can be written as follows: \(8^{30}*2\)
now the remainder of the expression when divided by \(7\) is given by : \(Rof [ 8^{30}*2 ]/7\)
==> \(Rof [(8 *8*8*8...30times)*2]/7\)
=> \(ROf[ (1*1*1*...30times)*2]/7\)
=>\(Rof[2]/7\)
==> remainder of the expression = 2
to understand above concept please follow following links :
https://gmatclub.com/forum/tips-how-to-get-remainders-92928.html
https://gmatclub.com/forum/numbers-86325-20.html#p715927
hope this will help.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
